We have \(\displaystyle \theta(\zeta(x,t))\).
The first derivative will be
\(\displaystyle \frac{d\theta}{d\zeta} = \frac{\partial \theta}{\partial x} \frac{dx}{d\zeta} + \frac{\partial \theta}{\partial t} \frac{dt}{d\zeta} = \frac{\partial \theta}{\partial x} - \frac{\partial \theta}{\partial t} \frac{1}{a}\)
The second derivative will be
\(\displaystyle \frac{d^2\theta}{d\zeta^2} = \frac{\partial}{\partial x}\left(\frac{\partial \theta}{\partial x} - \frac{\partial \theta}{\partial t} \frac{1}{a}\right)\frac{dx}{d\zeta} + \frac{\partial}{\partial t}\left(\frac{\partial \theta}{\partial x} - \frac{\partial \theta}{\partial t} \frac{1}{a}\right)\frac{dt}{d\zeta}\)
\(\displaystyle = \frac{\partial^2 \theta}{\partial x^2} - \frac{\partial^2 \theta}{\partial x\partial t} \frac{1}{a} - \frac{\partial^2 \theta}{\partial t\partial x}\frac{1}{a} + \frac{\partial^2 \theta}{\partial t^2} \frac{1}{a^2}\)
\(\displaystyle = \frac{\partial^2 \theta}{\partial x^2} - \frac{\partial^2 \theta}{\partial x\partial t}\frac{2}{a} + \frac{\partial^2 \theta}{\partial t^2} \frac{1}{a^2}\)
Substituting this in the original DE, you will get
\(\displaystyle \frac{\partial^2 \theta}{\partial x^2} - \frac{\partial^2 \theta}{\partial x\partial t}\frac{2}{a} + \frac{\partial^2 \theta}{\partial t^2} \frac{1}{a^2} = c_1\theta + c_2\theta^3\)